Edexcel IAL Maths
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Evaluating definite integrals of sums of powers of x, including roots, fractions and unknown limits
Why there is no + c in a definite integral
The area between a curve and the x-axis
What happens when part of the region is below the axis
Areas of regions bounded by a curve and a straight line, or between two curves
The trapezium rule and the effect of using more strips
Whether the trapezium rule over- or underestimates, and percentage error
1. A definite integral ∫ₐᵇ f(x) dx = F(b) - F(a), where F is any integral of f, is a number, and it equals the area under the curve between x = a and x = b as long as the curve stays above the x-axis there.
2. The constant c cancels, so leave it out of definite integrals.
3. Below the axis the integral is negative. If the curve crosses the axis, find where, integrate each part separately and add the sizes.
4. Between two graphs: area = ∫ₐᵇ (top - bottom) dx, with a and b where they meet. Composite regions: add or subtract simple pieces such as triangles.
5. Trapezium rule: ∫ₐᵇ y dx ≈½ h{y₀ + yₙ + 2(y₁ + y₂ + ⋯+ yₙ₋₁)}, h = (b-a)/n. Curve bending up: overestimate. Curve bending down: underestimate. More strips: smaller error.
In P1, integrating gave a family of curves: ∫2x dx = x² + c, one curve for each value of c. Put two numbers on the integral sign and the answer becomes a single number instead, and that number turns out to be an area.
Definite integral.
An integral with limits,
∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) - F(a), where F′(x) = f(x).
Its value is a number. a is the lower limit and b the upper limit.
Integrate as in P1, write the result in square brackets with the limits attached, then substitute: upper value minus lower value.
The constant of integration cancels: (F(b) + c) - (F(a) + c) = F(b) - F(a). That is why there is no + c in a definite integral.
For example, ∫₁³ (3x² - 4x + 1) dx = [x³ - 2x² + x]₁³ = (27 - 18 + 3) - (1 - 2 + 1) = 12 - 0 = 12.
Swapping the limits changes the sign: ∫₃¹ (3x² - 4x + 1) dx = -12.
WORKED EXAMPLE · QUESTION
Find the exact value of ∫₁⁴ (2√x - 3/x²) dx.
SOLUTION
Write each term as a power of x: 2x^(1/2) - 3x⁻².
Integrate (raise each power by one, divide by the new power):
[(4/3) x^(3/2) + 3x⁻¹]₁⁴ = [(4/3) x^(3/2) + 3/x]₁⁴
Substitute the limits, using 4^(3/2) = 8: ((32/3) + ¾) - ((4/3) + 3) = (28/3) - (9/4).
Combine over a common denominator: (112/12) - (27/12). (85/12)
"Exact" means leave it as a fraction; the decimal 7.0833… would not be in the form asked.
WORKED EXAMPLE · QUESTION
Given that k > 1 and ∫₁ᵏ (4x - 1) dx = 14, find the value of k.
SOLUTION
Integrate and substitute the limits; the upper limit is k: [2x² - x]₁ᵏ = (2k² - k) - (2 - 1) = 2k² - k - 1.
Set equal to 14 and solve the quadratic: 2k² - k - 15 = 0 ⇒(k - 3)(2k + 5) = 0.
Use the condition k > 1: k = -(5/2) is rejected. k = 3
Common error
Losing a sign when subtracting the lower value. Put brackets round both F(b) and F(a) before subtracting: (…) - (…). The lower value often contains negatives, and every one of them changes sign.
Why should an integral give an area? Think of the area under the curve from a up to a moving right-hand edge x, and call it A(x). Moving the edge a tiny step δx adds a thin strip of height about y, so δA ≈y δx, and dA/dx = y. The area function is therefore an integral of y, and the area between a and b is A(b) - A(a): exactly the definite integral.
Area under a curve
(not in the formulae booklet: learn it). For y = f(x) ≥0 when a ≤x ≤b, the region bounded by the curve, the x-axis and the lines x = a and x = b has
Area = ∫ₐᵇ y dx.
If the region is described as "bounded by the curve and the x-axis", the limits are where the curve meets the axis: solve f(x) = 0 first.
WORKED EXAMPLE · QUESTION
The curve C has equation y = 6x - x² - 5. The region R is bounded by C and the x-axis. Find the exact area of R.
SOLUTION
Find the limits, where the curve meets the x-axis: 6x - x² - 5 = 0 ⇒x² - 6x + 5 = 0 ⇒(x - 1)(x - 5) = 0, so x = 1 and x = 5.
Integrate: ∫₁⁵ (6x - x² - 5) dx = [3x² - x³/3 - 5x]₁⁵.
Substitute the limits: (75 - (125/3) - 25) - (3 - ⅓ - 5) = (25/3) - (-(7/3)). (32/3)
The answer is positive, as it must be for a region above the axis: a quick check worth making every time.
Exam tip
Sketch the curve, even roughly, before integrating for an area. The sketch tells you the limits, whether the region is above or below the axis, and whether you need to add or subtract another shape.
Where the curve is below the x-axis, y < 0, so each thin strip y δx is negative and the definite integral there is negative. The area is the size (modulus) of that integral.
If the curve crosses the axis between the limits, the positive and negative parts cancel. For example ∫₋₂² x³ dx = 0, although the region between y = x³ and the axis from x = -2 to 2 has area 8.
The method is therefore: find where the curve meets the axis, integrate each part separately, make each answer positive, then add.
WORKED EXAMPLE · QUESTION
The curve y = x(x - 1)(x - 3) meets the x-axis at x = 0, 1 and 3. (a) Evaluate ∫₀³ x(x - 1)(x - 3) dx. (b) Find the total area of the regions bounded by the curve and the x-axis for 0 ≤x ≤3.
SOLUTION
Expand, then integrate: x(x - 1)(x - 3) = x³ - 4x² + 3x, so F(x) = x⁴/4 - (4/3) x³ + (3/2) x².
(a) One integral from 0 to 3: F(3) - F(0) = ((81/4) - 36 + (27/2)) - 0. -(9/4)
(b) Split at x = 1, where the curve crosses the axis: ∫₀¹ = F(1) - F(0) = ¼ - (4/3) + (3/2) = (5/12); ∫₁³ = F(3) - F(1) = -(9/4) - (5/12) = -(8/3). Add the sizes: (5/12) + (8/3). Area = (37/12)
The answer to (a), -(9/4), is the signed total (5/12) - (8/3); it is not an area.
Common error
Giving a negative number as an area, or integrating straight across a root. An area is never negative, and an integral across a point where the curve crosses the axis gives the difference of the two areas, not their sum.
Two approaches cover almost every region you will meet, and both integrate with respect to x.
Split into pieces. A region bounded by a curve, a straight line and the x-axis is often "area under the curve" plus (or minus) a triangle or trapezium. Use geometry for straight-edged pieces: it is quicker than integrating.
Top minus bottom. Between two graphs y = f(x) (upper) and y = g(x) (lower) meeting at x = a and x = b, the area is ∫ₐᵇ (f(x) - g(x)) dx. This works even if part of the region is below the x-axis, because f - g stays positive.
Find the x-coordinates where the graphs meet by solving f(x) = g(x); these are the limits.
WORKED EXAMPLE · QUESTION
The curve y = 2x², x ≥0, and the line y = 12 - 2x meet at the point P. The region R is bounded by the curve, the line and the x-axis. (a) Find the coordinates of P. (b) Find the exact area of R.
SOLUTION
(a) Solve simultaneously: 2x² = 12 - 2x ⇒x² + x - 6 = 0 ⇒(x + 3)(x - 2) = 0; x ≥0, so x = 2. P(2, 8)
(b) Split at x = 2. Under the curve from 0 to 2: ∫₀² 2x² dx = [⅔ x³]₀² = (16/3). Under the line from 2 to where it meets the axis (12 - 2x = 0 ⇒x = 6) is a triangle: ½ ×(6 - 2) ×8 = 16. Add: (16/3) + 16. Area of R = (64/3)
WORKED EXAMPLE · QUESTION
The curve C has equation y = x(x - 3)² and the line L has equation y = x. (a) Show that C and L meet where x = 0, x = 2 and x = 4. (b) Show that the two finite regions enclosed by C and L have equal area, and find that area.
SOLUTION
(a) Set the equations equal: x³ - 6x² + 9x = x ⇒x(x² - 6x + 8) = 0 ⇒x(x - 2)(x - 4) = 0, so x = 0, 2, 4 ✓
(b) Decide which graph is on top in each interval, using a test value. At x = 1: C gives 4, L gives 1, so C is above. At x = 3: C gives 0, L gives 3, so L is above.
Top minus bottom on [0, 2], with C - L = x³ - 6x² + 8x: ∫₀² (x³ - 6x² + 8x) dx = [x⁴/4 - 2x³ + 4x²]₀² = (4 - 16 + 16) - 0 = 4.
Top minus bottom on [2, 4] is L - C, the negative of the same expression: -[x⁴/4 - 2x³ + 4x²]₂⁴ = -((64 - 128 + 64) - 4) = 4. Both regions have area 4, so they are equal
∫₀⁴ (C - L) dx = 0: integrating across the crossing point at x = 2 cancels the two regions exactly, the same trap as for regions below the axis.
Exam tip
Write "area = ∫ₐᵇ (top - bottom) dx" with the actual expressions before simplifying. Subtracting the line from the curve first, then integrating once, is quicker and safer than finding two areas under graphs and subtracting.
Many functions cannot be integrated with the rules you know at this stage: 2ˣ, √(1 + x³), log₁₀ x. Data from an experiment has no formula at all. The trapezium rule replaces the curve by straight chords between equally spaced points and adds the areas of the trapezia underneath.
The trapezium rule
(in the formulae booklet: know how to use it):
∫ₐᵇ y dx ≈½ h{y₀ + yₙ + 2(y₁ + y₂ + ⋯+ yₙ₋₁)}
n = number of strips; h = (b - a)/n = width of each strip; yᵣ = f(a + rh) are the ordinates, and there are n + 1 of them.
In words: half the strip width × (first + last + twice everything in between). Each trapezium has area ½ h(yᵣ + yᵣ₊₁); each inner ordinate is a side of two trapezia, which is why it counts twice.
WORKED EXAMPLE · QUESTION
Use the trapezium rule with 4 strips to estimate ∫₀² 2ˣ dx. Give the ordinates to 4 decimal places and your answer to 4 significant figures.
SOLUTION
Strip width and the x-values: h = (2 - 0)/4 = 0.5; x = 0, 0.5, 1, 1.5, 2.
| x | 0 | 0.5 | 1 | 1.5 | 2 |
| y = 2ˣ | 1 | 1.4142 | 2 | 2.8284 | 4 |
Apply the rule, end ordinates once and inner ordinates twice: ½ ×0.5 ×{(1 + 4) + 2(1.4142 + 2 + 2.8284)} = 0.25 ×{5 + 12.4852} = 0.25 ×17.4852. ≈4.371
Keep the ordinates to at least 4 decimal places in the working; rounding them to 2 d.p. can change the final figure.
Common error
Mixing up strips and ordinates. Four strips need five x-values; h = (b - a)/n uses the number of strips, not the number of ordinates.
If the curve bends upwards (convex, like y = x² or 2ˣ) across the interval, every chord lies above the curve, so the trapezium rule gives an overestimate.
If the curve bends downwards (concave, like y = √x), every chord lies below the curve: an underestimate.
If the curve changes from one to the other inside the interval, the errors partly cancel and you cannot say in general.
More strips (smaller h) make the chords hug the curve more closely, so the estimate improves. For 2ˣ on [0, 2], 2, 4 and 8 strips give 4.5, 4.371 and 4.339, closing in on the true value 4.328 (to 4 s.f.) from above.
Percentage error
(not in the formulae booklet: learn it):
percentage error = (|estimate - exact value|)/(exact value) ×100%
You can only find it when the exact value is available, usually because the same integral can also be found by integration.
WORKED EXAMPLE · QUESTION
(a) Use the trapezium rule with 4 strips to estimate ∫₁³ 6/x² dx. (b) State, with a reason, whether your answer is an overestimate or an underestimate. (c) Find the exact value of ∫₁³ 6/x² dx, and hence the percentage error in your estimate.
| Strips n | Width h | Estimate | Percentage error |
|---|---|---|---|
| 2 | 1 | 4.833 | 20.8% |
| 4 | 0.5 | 4.23 | 5.75% |
| 8 | 0.25 | 4.059 | 1.49% |
The same integral with 2, 4 and 8 strips: each halving of h cuts the error to about a quarter.
SOLUTION
(a) h = (3 - 1)/4 = 0.5.
| x | 1 | 1.5 | 2 | 2.5 | 3 |
| y = 6/x² | 6 | 2.6667 | 1.5 | 0.96 | 0.6667 |
0.25 ×{(6 + 0.6667) + 2(2.6667 + 1.5 + 0.96)} = 0.25 ×{6.6667 + 10.2534}. ≈4.23
(b) y = 6x⁻² bends upwards on [1, 3] (it is decreasing ever more slowly), so each chord lies above the curve. Overestimate
(c) ∫₁³ 6x⁻² dx = [-6/x]₁³ = -2 - (-6) = 4. Percentage error = (4.23 - 4)/4 ×100%. 5.75%
The exact value is smaller than the estimate, consistent with part (b). Use the unrounded estimate (here it happens to be exactly 4.23).
Exam tip
"State, with a reason, whether … overestimate or underestimate" needs both parts: the shape (curve bends up or down, chords above or below the curve) and the conclusion. A sketch with one strip and its chord is a good reason.
Exam tip 1 — show the substitution
Write out F(b) - F(a) with both brackets before simplifying. A calculator can evaluate a definite integral, but the working is where the method is shown, and "use algebraic integration" means a calculator value alone is not enough.
Exam tip 2 — limits from the question, not from habit
Limits come from the region: where the curve meets the axis, where the graphs meet, or the given lines x = a and x = b. Solve for them first and mark them on a sketch.
Exam tip 3 — areas are positive
If an integral for an area comes out negative, the region is below the axis (or you used bottom − top). State the area as a positive value, and split at every crossing point.
Exam tip 4 — trapezium rule layout
Give h, the table of ordinates and the bracket {y₀ + yₙ + 2(…)} as separate lines. Keep 4 or more decimal places in the table and round only the final answer, to the accuracy asked.
Exam tip 5 — "hence" with the trapezium rule
If ∫ₐᵇ f(x) dx ≈T, then ∫ₐᵇ (f(x) + k) dx ≈T + k(b - a) and ∫ₐᵇ mf(x) dx ≈mT. Use your earlier answer rather than starting a new table.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A definite integral needs + c. | The constant appears in F(b) and F(a) and cancels. | Leave + c out of definite integrals. |
| A definite integral always equals an area. | Parts below the x-axis count as negative. | Split at the roots and add the sizes of the parts. |
| An integral of zero means there is no region. | Equal areas above and below the axis cancel. | Check where the curve crosses the axis between the limits. |
| Area between two graphs = area under the top one. | That includes the region under the lower graph too. | Integrate (top − bottom) between the meeting points. |
| With n strips there are n ordinates. | Each strip has two edges, shared with its neighbours. | n strips need n + 1 ordinates; h = (b-a)/n. |
| The trapezium rule always overestimates. | That depends on which way the curve bends. | Bends up: overestimate. Bends down: underestimate. |
Integration (P1) and further integration (P3, P4)
Every definite integral starts with the indefinite integral from P1: rewriting roots and fractions as powers, then raising each power by one. Definite integrals of eˣ, 1/x and trigonometric functions come in P3; in P4 they give volumes of revolution and are combined with substitution and integration by parts. The trapezium rule returns for functions that still cannot be integrated exactly.
Variable acceleration (M2)
The area under a velocity–time graph is the displacement, ∫_(t₁)^(t₂) v dt. When the velocity changes sign, total distance needs the same splitting as for regions below the axis.
IAL Physics
Areas under graphs carry physical meaning: displacement from a velocity–time graph, work done from a force–extension graph, impulse from a force–time graph. When the graph comes from data, a trapezium estimate is the natural method.
| Formula | Use | In the formulae list? |
|---|---|---|
| ∫ₐᵇ f(x) dx = F(b) - F(a), F′ = f | evaluating a definite integral | No — learn it |
| ∫xⁿ dx = xⁿ⁺¹/(n+1) + c, n ≠-1 | integrating each term before substituting limits | No — learn it |
| Area = ∫ₐᵇ y dx (curve above the axis) | area between a curve and the x-axis | No — learn it |
| Area = ∫ₐᵇ (yₜₒₚ - y_bottom) dx | area between two graphs meeting at x = a, b | No — derive it |
| ∫ₐᵇ y dx ≈½ h{y₀ + yₙ + 2(y₁ + y₂ + ⋯+ yₙ₋₁)}, h = (b-a)/n | estimating a definite integral with n strips | Yes |
| percentage error = (|estimate - exact|)/exact ×100% | judging the accuracy of an estimate | No — learn it |
1. Evaluate ∫₀² (x³ + 1) dx.
2. Find the exact value of ∫₁⁴ √x dx.
3. ∫₋₁¹ x³ dx = 0. Does that mean the curve y = x³ encloses no area with the x-axis between −1 and 1?
4. How many ordinates are needed to use the trapezium rule with 6 strips on 0 ≤x ≤3, and what is h?
5. The curve y = √(1 + x) bends downwards for x ≥0. Does the trapezium rule over- or underestimate ∫₀³ √(1 + x) dx?
6. Given that ∫₀² 2ˣ dx ≈4.371 (Worked example 7), estimate ∫₀² (2ˣ + 3) dx without a new table.
1. [x⁴/4 + x]₀² = (4 + 2) - 0 = 6.
2. [⅔ x^(3/2)]₁⁴ = ⅔(8 - 1) = (14/3).
3. No. The part from −1 to 0 is below the axis and gives -¼; the part from 0 to 1 gives +¼. They cancel in the integral, but the total area is ½.
4. 7 ordinates (x = 0, 0.5, 1, …, 3), with h = (3/6) = 0.5.
5. Underestimate: the curve bends downwards, so every chord lies below it and each trapezium misses a sliver of area.
6. Adding 3 adds a rectangle of height 3 and width 2: 4.371 + 3 ×2 = 10.371 ≈10.37.
Check understanding
The answer and each step appear one click at a time.
The table shows some values of y = √(x³ + 1), each given to 4 decimal places where appropriate.
Figure: the curve with the four trapezium strips
(a) Complete the table, giving the missing value of y to 4 decimal places. [1]
(b) Use the trapezium rule with all the values in the table to find an estimate for ∫₀² √(x³ + 1) dx, giving your answer to 3 significant figures. [4]
(c) State how the trapezium rule could be used to obtain a more accurate estimate for this integral. [1]
| x | 0 | 0.5 | 1 | 1.5 | 2 |
| y | 1 | 1.0607 | 2.0917 | 3 |
Mark scheme click to reveal each step
at x = 1: √2 = 1.4142
strip width h = 0.5
correct structure: ½h{(first + last) + 2(sum of the others)}
½ × 0.5 × {(1 + 3) + 2(1.0607 + 1.4142 + 2.0917)} = 0.25 × 13.1332 ft from their table value
3.28
use more strips (a smaller strip width h), e.g. 8 strips of width 0.25
The integral I = ∫₀² 3ˣ dx is to be estimated numerically, because 3ˣ cannot be integrated with the methods of this unit. Values of 3ˣ that are not exact are rounded to 4 decimal places.
Figure: y = 3ˣ for 0 ≤ x ≤ 2 with the ordinates used in the trapezium rule
(a) Complete the table. [1]
(b) Use the trapezium rule with all the values in the completed table to find an estimate for ∫₀² 3ˣ dx, giving your answer to 3 significant figures. [3]
(c) State, with a reason, whether your answer to part (b) is an overestimate or an underestimate. [2]
(d) Using your answer to part (b), find an estimate for ∫₀² (3ˣ⁺¹ + 4x) dx. [2]
| x | 0 | 0.5 | 1 | 1.5 | 2 |
| 3ˣ | 1 | 3 | 9 |
Mark scheme click to reveal each step
1.7321 at x = 0.5 and 5.1962 at x = 1.5, both correct
h = 0.5
½ × 0.5 × {(1 + 9) + 2(1.7321 + 3 + 5.1962)}
7.46
the curve y = 3ˣ bends upwards (convex) on 0 ≤ x ≤ 2, so each chord lies above the curve
so it is an overestimate (dependent on the reason)
3ˣ⁺¹ = 3 × 3ˣ, so the estimate is 3 × 7.46… + [2x²]₀² = 22.39… + 8
30.4 ft from their (b)
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ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only.
Pearson Edexcel and Cambridge are trademarks of their owners; ExaminerPrep is not affiliated with any exam board.